Start with the current, not the wattage
This calculator needs the inverter’s maximum continuous DC input current, which is not the same number as its AC output rating.
If the datasheet lists it, use that. Otherwise:
I(DC) = continuous AC watts ÷ inverter efficiency ÷ lowest battery voltage under load
Two details matter. Use efficiency, typically 0.88 to 0.94 — the inverter draws more DC power than it delivers as AC. And use the lowest battery voltage you expect under load, not the nominal label. A 24 V bank under a heavy draw sits nearer 23 V than 24 V, and that difference alone moves the current by about 4 %.
Worked: a 3000 W inverter at 90 % efficiency on a 24 V bank sagging to 23 V draws 3000 ÷ 0.9 ÷ 23 ≈ 145 A.
Inverter cable current by system voltage
Continuous DC input current for common inverter sizes at 90 % efficiency and nominal bank voltage. This is the number to type into the calculator above.
| Inverter | 12 V bank | 24 V bank | 48 V bank |
|---|---|---|---|
| 1000 W | 93 A | 46 A | 23 A |
| 1500 W | 139 A | 69 A | 35 A |
| 2000 W | 185 A | 93 A | 46 A |
| 3000 W | 278 A | 139 A | 69 A |
| 5000 W | 463 A | 231 A | 116 A |
At 12 V, a 5000 W inverter needs over 460 A. That is beyond any single conductor in the NEC tables up to 500 kcmil and requires parallel conductors — the clearest possible argument for building larger systems at 48 V.
What actually fails on this circuit
The termination, not the cable. At 150 A a loose or under-crimped lug generates heat inside a connection you cannot inspect. Use a hydraulic crimper and listed lugs, and check the terminal temperature rating — most inverter and battery terminals are listed for 75 °C, which caps the ampacity column you may use under NEC 110.14(C) regardless of your conductor’s 90 °C jacket.
A smaller negative cable. The same current returns through it. The voltage drop figure above already accounts for both conductors.
Protection at the wrong end. The fuse belongs at the battery, as close to the positive terminal as practical, because the battery is the source of fault current.
Cable draped across a hot surface. The ampacity table assumes 30 °C ambient. An engine bay or a sealed battery box in summer is not 30 °C, and the calculator’s ambient field exists for exactly this.
Where this sits in the system
Sizing this cable gives you the current the battery-side overcurrent device has to be built around, and it is often the number that reveals whether your system voltage choice was right. If the answer here is uncomfortable, the fix is usually a higher bank voltage rather than more copper.